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What exactly is centrifugal force 
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#55
Jan3013, 08:49 AM

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Just throwing in my 2 cents worth.



#56
Jan3013, 09:01 AM

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#57
Jan3013, 09:03 AM

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#58
Jan3013, 09:10 AM

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#59
Jan3013, 09:20 AM

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#60
Jan3013, 09:23 AM

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#61
Jan3013, 09:25 AM

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#62
Jan3013, 09:43 AM

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#63
Jan3013, 09:48 AM

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"If one uses noninertial, curvilinear coordinates, then additional terms appear in the equations of motion." One sentence. Everything else is an argument for a particular way of looking at those additional terms. 


#64
Jan3013, 09:52 AM

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#65
Jan3013, 09:55 AM

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#66
Jan3013, 09:59 AM

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(1) They are vector quantities, meaning that they exist in EVERY coordinate system. The components change when you change coordinate systems, but a vector is a geometric quantity that is independent of coordinates. (2) Real forces have corresponding reaction forces, leading to conservation of momentum. In contrast, "inertial forces" are artifacts of a particular choice of coordinates. They don't have corresponding reaction forces. They can be made to disappear by choosing the appropriate coordinate system. 


#67
Jan3013, 10:11 AM

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But all of that is not relevant to the "centrifugal"label for the reason I state in post #64. 


#68
Jan3013, 10:13 AM

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#69
Jan3013, 10:29 AM

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I suppose I have to take your point because my suggestion is not falsifiable. But I think that the inverse,  i.e. that Science definitely can establish 'real truth' probably is falsifiable. So far, we have found this as our experience has been that Science, and its models, continuously changes to fit new evidence. It has to be true that Science endeavors to avoid saying what things 'really are' because there are so many examples of two or more, equally valid 'realities'. (Note, I write "Science" and not 'Scientists'  who are human and fallible and seldom view things without the distraction of some sort of faith). 


#70
Jan3013, 10:33 AM

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AM 


#71
Jan3013, 10:44 AM

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#72
Jan3013, 10:45 AM

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[itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex] The lefthand side is a fact about the motion of the object, and the righthand side is about the external situation affecting that motion. In terms of coordinates: [itex](\frac{\stackrel{\rightarrow}{dU}}{dt})^i = \frac{dU^i}{dt} +[/itex] sum over [itex]j, k[/itex] of [itex] \Gamma^i_{jk} U^j U^k[/itex] where [itex]\Gamma^i_{jk}[/itex] are the socalled "connection coefficients", which are due to using nonconstant basis vectors. So the full equations of motion, in terms of components, are: [itex]m(\frac{dU^i}{dt} +[/itex] sum over [itex]j, k[/itex] of [itex] \Gamma^i_{jk} U^j U^k) = F^i[/itex] What the idea of "fictitious forces" amounts to is moving the extra terms from the left side (where they describe motion) to the right side (where they are treated as forces): [itex]m \frac{dU^i}{dt} = F^i + F_{inertial}^i[/itex] where [itex]F_{inertial}^i =  m[/itex] sum over [itex]j, k[/itex] of [itex] \Gamma^i_{jk} U^j U^k[/itex] What difference does it make whether you group it on the left side, or the right side? Well, for one thing, when it comes to figuring out the reaction forces (Newton's third law), only the [itex]F^i[/itex] term is relevant. There are no reaction forces to [itex]F_{inertial}^i[/itex]. For another, since real forces are vectors, the components transform in a standard way under a coordinate change: If you change coordinates from [itex]x^i[/itex] to [itex]y^b[/itex], then [itex]F^b = [/itex] sum over [itex]i[/itex] of [itex]\dfrac{\partial y^b}{\partial x^i} F^i[/itex] "Inertial forces" DON'T transform that way. So sure, you can group whatever terms together you want, and call them whatever you want to call them, but when it comes to reasoning about the physics, you have to separate out the "real" forces from the "inertial" forces. You're basically doing extra steps that have to be undone later. 


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